Repair of Appendix D for m = 0 error REVEIWED

Repair of Appendix D for m = 0 error PhL 3.6.14 Just as it says, I found an error ONLY in the Appendix D results for m = 0. I guess this error did not affect the comparison with Chapter 2 or I would have found it earlier. I think is all OK. I luckily found an error today and will now attempt to fix it. Here first is a review of the math with Maple verification. I start off with First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - βd2 (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) and I want to apply these two boundary conditions, Er(r=a,m) = (jω/σ) Nm (D.2.26) Eφ(r=a,m) = 0 (D.2.27) The boundary conditions then say, in terms of the last two equations in (D.2.21), am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 . (2) OK I now enter these equations into Maple and have it solve for am (= am_s) and Km ( = Km_s), Then I enter my solutions and see if they are the same, The only other thing to check is the sum thing: So I am convinced that my general results are correct as stated in (D.2.28). am = (jω/2σ) 2m Nm . => a0 = 0 (D.2.28) = (jω/2σ) Nm [ – ] => = (jω/σ) N0 // J-1(z) = - J1(z) (+ ) = (jω/2σ) Nm [ + ] and I rewrite the first one as = (jω/2σ) 2 Nm . (5) The issue at hand is the above ratio when m = 0. We know that a0 = 0 from (D.2.28) and of course m = 0 so we have a 0/0 situation. The correct ratio is given by (5) above, = (jω/2σ) 2 N0 = - (jω/2σ) 2 N0 but I accidentally set this ratio to to zero in several places! So I keep marching through Appendix D now. I verified (visual tracking) all the algebra of inserting the coefficients(D.2.28) into (D.2.21) to get (D.2.33). That algebra luckily is written out above (D.2.33). Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] a = radius ηm ≡ Er(r,m) = (j/4) ηm I Rdc (aβd) [ + - ] x = β'r Eφ(r,m) = (1/4) ηm I Rdc (aβd) [ - + + ] xa = β'a My next collection however goes back to the coefficients, it is (D.4.9), and this is the box that I used to do the m = 0 reductions: Ez(r,m) = - j (β'/βd) Jm(x) x = β'r β'2 = β2 - βd2 Er(r,m) = am x-1 Jm(x) + Jm+1(x) . jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.21) Bz(r,m) = (β'/ω) ( + )Jm(x) Br(r,m) = j(β'/ω){ + ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } Bφ(r,m) = (β'/ω){ - ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } , (D.4.8) The key facts are: (+ )|m=0 = (jω/2σ) N0 [ + ] = (jω/2σ) N0 [ - ] = 0 = (jω/σ) N0 = Here then is my corrected (D.6.1) Summary of E and B fields inside a round wire ( m = 0 only ) (D.6.1) Ez(r,0) = - j (β'/βd) J0(x) // large a0 = 0 x = β'r β'2 = β2 - βd2 Er(r,0) = J1(x) . // small = (j/2) (aβd) I Rdc jEφ(r,0) = 0 // small Rdc = Bz(r,0) = 0 // small ~ β' Br(r,0) = 0 // very small ~ βd Bφ(r,0) = (β'/ω) ( + ) J1(x) // large ~ β' (β'/βd) Ouch, I have not one but three mistakes in this box. I did a long round of edits just now in lines doc, see edit log, and I included the above fixes.